What are the factors of 352?

1 Answer
Nov 5, 2016

#352 = 2^5*11# has positive integer factors:

#1, 2, 4, 8, 11, 16, 22, 32, 44, 88, 176, 352#

Explanation:

We can draw a factor tree for #352# like this:

#color(white)(0000)352#
#color(white)(000)"/"color(white)(000)"\"#
#color(white)(00)2color(white)(0000)176#
#color(white)(000000)"/"color(white)(000)"\"#
#color(white)(00000)2color(white)(0000)88#
#color(white)(000000000)"/"color(white)(00)"\"#
#color(white)(00000000)2color(white)(000)44#
#color(white)(00000000000)"/"color(white)(00)"\"#
#color(white)(0000000000)2color(white)(000)22#
#color(white)(0000000000000)"/"color(white)(00)"\"#
#color(white)(000000000000)2color(white)(000)11#

We notice that #352# ends with an even digit, so can be divided by #2# to give #176#. Then #176# is even again, etc. Finally we get to #11# which is a prime number.

So the prime factorisation of #352# is:

#352 = 2^5*11#

We can list all positive integer factors of #352# by listing all of the powers of #2# from #2^0 = 1# up to #2^5 = 32# and the same multiplied by #11#:

#1, 2, 4, 8, 16, 32, 11, 22, 44, 88, 176, 352#

or in ascending order:

#1, 2, 4, 8, 11, 16, 22, 32, 44, 88, 176, 352#